Magnetic field in a cylindrical region is varying as B = kt/r, ('k' is a positive constant, 't' is time and 'r' is distance from the center). B is directed upwards perpendicular to the cylindrical surface. At the origin (r=0), magnetic field is negligibly small. A charge particle'q' of mass 'm' is released from rest at a distance d from the center. What's the force on the charged particles just after the release?

Since, the charge particle was just released, so it'd have 0 velocity so magnetic force won't act on it directly. So I tried to find the force using the electric field induced by the varying magnetic field.I got the electric field as k/2 (using Maxwell's equation) and so the force as kq/2. But the answer in my textbook is given as kq. Would someone please help me understand where am I getting wrong?

  • $\begingroup$ what is the direction of B? $\endgroup$ Jun 10, 2022 at 14:26
  • $\begingroup$ Even though this is homework, .show your steps before asking questions like this . With enough assumptions about the direction of the B field, ( being upwards, and cylindrical symmetry) the initial force is as your textbook has stated, with the force directed clockwise. This is until it actually starts moving, and then we must consider the magnetic force $\endgroup$ Jun 10, 2022 at 14:29
  • $\begingroup$ If $\vec{B}$ (and $\hat z$) is upwards. $\vec{F}= -kq \hat \phi$ $\endgroup$ Jun 10, 2022 at 14:35
  • $\begingroup$ Sorry, I didn't mention in the question. B is upwards $\endgroup$
    – Aspirant
    Jun 10, 2022 at 17:18

1 Answer 1


Your confusion comes from your miscalculation of the flux.

$$\iint \vec{B} \cdot \vec{da}≠ |\vec{B}| \iint |\vec{da}|$$

Except where $|\vec{B}|$ is a constant,and in the same direction as $\vec{da}$

$$\iint \vec{B} \cdot \vec{da}= \int_{0}^{2\pi} \int_{0}^{R} [\frac{kt}{r}] r dr d\phi$$

$\vec{F} = -kq \hat \phi$ for an upward $\vec{B}$


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